heating an ore of antimony and iron II sulfide.
Sb2S3+3Fe-->2Sb+3FeS
when 15g Sb2S3 reacts with an excess of Fe, 9.84g Sb is produced. what is the pecent yeild of this reaction?
Convert 15 g Sb2S3 to mols using mols = g/molar mass.
Convert mols Sb2S3 to mols mols Sb using the coefficients in the balanced equation.
Convert mols Sb to grams using g = mols x molar mass. This is the theoretical yield.
%yield = [(actual yield)/(theoretical yield)]*100
actual yield is 9.84 g.
theoretical yield is calculated above.
To calculate the percent yield of a reaction, we need to compare the actual yield with the theoretical yield.
First, let's calculate the molar mass of Sb2S3:
Sb2S3: (2 * 121.75 g/mol) + (3 * 32.07 g/mol) = 339.69 g/mol
Next, we can calculate the theoretical yield of Sb based on the balanced equation:
1 mole of Sb2S3 produces 2 moles of Sb.
So, the theoretical yield of Sb can be calculated as follows:
(15 g Sb2S3) * (1 mol Sb2S3 / 339.69 g Sb2S3) * (2 mol Sb / 1 mol Sb2S3) * (121.75 g Sb / 1 mol Sb) = 27.28 g Sb
Now, we can calculate the percent yield using the formula:
Percent Yield = (Actual Yield / Theoretical Yield) * 100
Plugging in the values gives us:
Percent Yield = (9.84 g / 27.28 g) * 100 ≈ 36.04%
Therefore, the percent yield of this reaction is approximately 36.04%.
To calculate the percent yield of a reaction, you need to compare the actual yield (the amount of Sb obtained in this case) with the theoretical yield (the amount of Sb that would be obtained if the reaction went to completion).
1. Start by calculating the theoretical yield of Sb.
The balanced equation tells us that the reaction produces 2 moles of Sb for every 1 mole of Sb2S3 reacted. To convert grams to moles, you need to know the molar mass of Sb2S3 and Sb.
- Molar mass of Sb2S3 = (121.76 g/mol x 2) + (32.06 g/mol x 3) = 339.64 g/mol
- Moles of Sb2S3 = 15g / 339.64 g/mol = 0.0441 mol
- Moles of Sb (theoretical yield) = 2 x 0.0441 mol = 0.0882 mol
2. Calculate the percent yield.
Percent yield = (actual yield / theoretical yield) x 100
Since the question provided that 9.84g of Sb is produced, we can calculate the actual yield in moles:
- Moles of Sb (actual yield) = 9.84g / 121.76 g/mol = 0.0807 mol
Now, plug the values into the formula:
Percent yield = (0.0807 mol / 0.0882 mol) x 100 = 91.6%
Therefore, the percent yield of this reaction is approximately 91.6%.